Chalk this one up to your book. I didn't read the question very well; in fact, I had to read it 3-4 times today before I caught why we were having trouble. The book answer is correct. 2.65 g Na2CO3 will make 250 mL of a 0.1M solution of Na2CO3. However, that isn't what the question asked. The question was to prepare 0.250L of a 0.1M aqueous solution of SODIUM IONS. So we have 2.65 grams Na2CO3 to prepare 250 mL of a 0.1 M Na2CO3 soln, and we want 0.1 M Na^+ (and there are two Na^+ in 1 molecule Na2CO3) so we need to divide our 2.65 g Na2CO3 by 2 to get where we want to go. Sorry about that. I don't always read these problems carefully enough. Thanks for bringing it back so we could get to the root of the problem.